The volume of O2 gas that is produced from 2.80 L of CO2 at STP is 1.40 L
write a balanced chemical equation
2Na2O2 + 2CO2 → 2Na2CO3 +O2
by use of mole ratio between CO2:O2 which is 2:1 the number of liters of O2 produced is therefore= 2.8 x 1/2 =1.40 Liters
Answer: The volume of [tex]O_2[/tex] produced will be 1.4 liters.
Explanation:
AT STP, the conditions are:
1 mole of a gas occupies 22.4 liters of volume.
The balanced chemical equation for the reaction follows:
[tex]2Na_2O_2+2CO_2\rightarrow 2Na_2CO_3+O_2[/tex]
By Stoichiometry of the reaction:
2 moles of carbon dioxide produces 1 mole of oxygen gas, which means that
44.8 liters of carbon dioxide gas produces 22.4 liters of oxygen gas.
So, 2.8 liters of carbon dioxide gas will produce = [tex]\frac{22.4}{44.8}\times 2.8=1.4liters[/tex] of oxygen gas.
Hence, the volume of [tex]O_2[/tex] produced will be 1.4 liters.
